Let's solve the problem step by step, as requested. ### Step 1: **Locate and classify all isolated singularities of \( g(z) \) and compute the residues at each singularity** #### **Locate singularities** The function is: \[ g(z) = \frac{e^{2z+1}}{(z^2 + 2z + 2)(z+1-i)} \] Set the denominator to zero to find singularities: - \( z^2 + 2z + 2 = 0 \implies z = -1 \pm i \) - \( z+1-i = 0 \implies z = -1+i \) Notice that \( z = -1+i \) is a root of both denominator factors, so it is a **double root** (order 2 pole). \( z = -1-i \) is a **simple root** (order 1 pole). #### **Classification** - \( z_1 = -1+i \): **Pole of order 2** (double pole) - \( z_2 = -1-i \): **Simple pole** (order 1) #### **Residue at \( z_2 = -1-i \)** For a simple pole: \[ \text{Res}(g, z_2) = \lim_{z \to z_2} (z - z_2)g(z) \] \[ = \lim_{z \to -1-i} \frac{e^{2z+1}}{(z+1+i)(z+1-i)} \] But since \( z+1-i \neq 0 \) at \( z = -1-i \), \[ = \frac{e^{2(-1-i)+1}}{((-1-i)+1+i)((-1-i)+1-i)} \] \[ = \frac{e^{-2-2i+1}}{(0)((-2i))} \] Wait, let's be careful with the algebra: - \( z+1+i \) at \( z = -1-i \): \( (-1-i) + 1 + i = 0 \) - \( z+1-i \) at \( z = -1-i \): \( (-1-i) + 1 - i = -2i \) So, \[ = \frac{e^{-1-2i}}{(0)(-2i)} \] But since the numerator is zero, it means the function has a removable singularity or a higher order pole at \( z = -1-i \), but since the denominator is quadratic and numerator is analytic, it's a simple pole. Let’s instead factor and simplify: \[ z^2 + 2z + 2 = (z+1+i)(z+1-i) \] So, \[ g(z) = \frac{e^{2z+1}}{(z+1+i)(z+1-i)(z+1-i)} = \frac{e^{2z+1}}{(z+1+i)(z+1-i)^2} \] So: - \( z=-1+i \): **Double pole** - \( z=-1-i \): **Simple pole** Residue at \( z_2 = -1-i \): \[ \text{Res}(g, -1-i) = \frac{e^{2(-1-i)+1}}{((-1-i)+1-i)^2} \] \[ = \frac{e^{-2-2i+1}}{(-2i)^2} \] \[ = \frac{e^{-1-2i}}{4(-1)} \] \[ = -\frac{e^{-1-2i}}{4} \] #### **Residue at \( z_1 = -1+i \) (Double Pole)** For a double pole: \[ \text{Res}(g, z_1) = \lim_{z \to z_1} \frac{d}{dz}\left[ (z-z_1)^2 g(z) \right] \] So, \[ (z-(-1+i))^2 g(z) = \frac{e^{2z+1}}{z+1+i} \] Differentiate with respect to \( z \), then set \( z = -1+i \): \[ \frac{d}{dz} \left( \frac{e^{2z+1}}{z+1+i} \right) = \frac{2e^{2z+1}(z+1+i) - e^{2z+1}}{(z+1+i)^2} \] \[ = \frac{e^{2z+1}(2(z+1+i) - 1)}{(z+1+i)^2} \] Now set \( z = -1+i \): \[ z+1+i = 2i \] \[ e^{2(-1+i)+1} = e^{-2+2i+1} = e^{-1+2i} \] \[ 2(z+1+i) - 1 = 2(2i) - 1 = 4i - 1 \] So, \[ \text{Res}(g, -1+i) = \frac{e^{-1+2i}(4i-1)}{(2i)^2} \] \[ = \frac{e^{-1+2i}(4i-1)}{4(-1)} \] \[ = -\frac{e^{-1+2i}(4i-1)}{4} \] --- ### Step 2: **Use Cauchy's Residue Theorem to compute the integral** #### **Check which singularities are inside \( C \)** Contour \( C = \{z \in \mathbb{C}: |z-i|=2\} \): - Center: \( i \) - Radius: 2 - \( z_1 = -1+i \): \( |-1+i - i| = |-1| = 1 < 2 \) (inside) - \( z_2 = -1-i \): \( |-1-i - i| = |-1-2i| = \sqrt{1+4} = \sqrt{5} \approx 2.24 > 2 \) (outside) So **only \( z_1 = -1+i \) is inside \( C \)**. #### **Apply Residue Theorem** \[ \int_C g(z)\,dz = 2\pi i \sum \text{Res}(g, \text{inside poles}) \] \[ = 2\pi i \left[ -\frac{e^{-1+2i}(4i-1)}{4} \right] \] \[ = -\frac{\pi i}{2} e^{-1+2i}(4i-1) \] --- ## **Final Answer** --- ### **Step 1:** - The function \( g(z) \) has a **double pole at \( z = -1+i \)** with residue \( -\frac{e^{-1+2i}(4i-1)}{4} \), and a **simple pole at \( z = -1-i \)** with residue \( -\frac{e^{-1-2i}}{4} \). ### **Step 2:** - Only the double pole at \( z = -1+i \) is inside the contour \( C \). By Cauchy's Residue Theorem, \[ \boxed{ \int_C g(z)\,dz = -\frac{\pi i}{2} e^{-1+2i}(4i-1) } \]